The x components of the average accelerations?

In summary: Or would you have to turn it into 21.904-9.36v1...or 21.904-9.36v2?Yes, you would have to turn it into 21.904-9.36v1...or 21.904-9.36v2.
  • #1
emily081715
208
4

Homework Statement


A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 12 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx. What are the x components of the average accelerations of the two chunks during the explosion?

Homework Equations


k=1/2mv^2

The Attempt at a Solution


k=1/2(8.5)(2.34)^2
= 23.27J

23.27+12=35.27J
35.27/0.16= 220.4m/s^2
i've never done a problem like this
 
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  • #2
What values are required to compute the average acceleration? ( or what's the formula for it).
 
  • #3
Aniruddha@94 said:
What values are required to compute the average acceleration? ( or what's the formula for it).
a=vf-vi/t
 
  • #4
emily081715 said:
a=vf-vi/t
Right. You've been given ##t## and the initial velocity for both the pieces. You need to find their final velocities. There are two unknowns, so we'll need 2 equations to solve them. Can you see what the equations can be?
 
  • #5
is one going to be the change in kinetic energy formula, and is the change momentum??
 
  • #6
emily081715 said:
is one going to be the change in kinetic energy formula, and is the change momentum??
is the initial velocity for both pieces just 2.34?
 
  • #7
emily081715 said:
is one going to be the change in kinetic energy formula, and is the change momentum??
Yes. You'll solve them to get the final velocities of both the pieces.
emily081715 said:
is the initial velocity for both pieces just 2.34?
That's right.
 
  • #8
Aniruddha@94 said:
Yes. You'll solve them to get the final velocities of both the pieces.

That's right.
for the mass do i use 8.5, or need to divide that by two since its split equally?
 
  • #9
emily081715 said:
for the mass do i use 8.5, or need to divide that by two since its split equally?
When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.
 
  • #10
Aniruddha@94 said:
When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.
but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?
 
  • #11
emily081715 said:
but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?
Yes, that's correct.
 
  • #12
Aniruddha@94 said:
Yes, that's correct.
i got answers of 20.48 and 25.16. they're wrong
 
  • #13
IMG_7666.JPG
 
  • #14
that was my work
 
  • #15
Your application of momentum conservation doesn't look right.
Suppose ##v## to be the initial velocity of the block and ##v_1## and ##v_2## be the final velocities of the two pieces. If the mass of the block is ##m##, the conservation principle gives $$p_i = p_f $$
$$ mv = \frac {m}{2}v_1 + \frac{m}{2}v_2 $$
This is your first equation. Similarly, for kinetic energy you'll get $$ \frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2 $$
Solve these two equations for ##v_1##and ##v_2##. Use them in the formula for average acceleration.
 
  • #16
Aniruddha@94 said:
Your application of momentum conservation doesn't look right.
Suppose ##v## to be the initial velocity of the block and ##v_1## and ##v_2## be the final velocities of the two pieces. If the mass of the block is ##m##, the conservation principle gives $$p_i = p_f $$
$$ mv = \frac {m}{2}v_1 + \frac{m}{2}v_2 $$
This is your first equation. Similarly, for kinetic energy you'll get $$ \frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2 $$
Solve these two equations for ##v_1##and ##v_2##. Use them in the formula for average acceleration.
i got answers of 74.56 and -74.31 for acceleration which are wrong
 
  • #17
emily081715 said:
i got answers of 74.56 and -74.31 for acceleration which are wrong
That doesn't match with my results either.
What values did you get for ##v_1## and ##v_2## ? Show your work.
 
  • #18
IMG_7667.JPG

i go values of 14.27 m/s and -9.59 m/s
 
  • #19
In the third line, you have substituted for ##v_2## in terms of ##v_1##. You seem to have made a mistake there.
From the first equation(momentum conservation), we get ##2v = v_1 + v_2##. Or ##v_2 = 4.68 - v_1## . Use this in your second equation.
 
  • #20
i got accelerations of -117.136 m/s^2 and 117.136m/s^2
 
  • #21
emily081715 said:
i got accelerations of -117.136 m/s^2 and 117.136m/s^2
is this correct?
 
  • #22
emily081715 said:
is this correct?
I don't think so. You're one step closer though. The average accelerations you get must have the same magnitude and opposite signs. Show your work for the above result.
 
  • #23
IMG_7669.JPG
 
  • #24
what did i miss?
 
  • #25
emily081715 said:
what did i miss?
In the third line you've written ##(4.68-v_1)^2## as ##( 21.9014-v^2_1)##. Are you allowed to do that?
 
  • #26
Aniruddha@94 said:
In the third line you've written ##(4.68-v_1)^2## as ##( 21.9014-v^2_1)##. Are you allowed to do that?
Wouldn't you apply the ^2 to both numbers?
 
  • #27
emily081715 said:
Wouldn't you apply the ^2 to both numbers?
Or would you have to turn it into 21.904-9.36v1 -v1^2
 
  • #28
The expansion for ##(a-b)^2## is ##(a^2+b^2-2ab)##
 
  • #29
That's what this is 21.904-9.36v1-v1^2
 
  • #30
i got accelerations now of 7.1 and -7.1 m/s^2
is this correct
 
  • #31
emily081715 said:
i got accelerations now of 7.1 and -7.1 m/s^2
is this correct
Still doesn't seem correct. What values of ##v_1## and ##v_2## did you get?
 
  • #32
1.204198 and 3.475
 
  • #33
emily081715 said:
1.204198 and 3.475
Those are not correct.

What do those give for the change in kinetic energy?
 
  • #34
emily081715 said:
Or would you have to turn it into 21.904-9.36v1 -v1^2
That last term should be positive:

21.904 - 9.36v1 + v12
 

Related to The x components of the average accelerations?

1. What are the x components of average acceleration?

The x components of average acceleration refer to the change in velocity in the horizontal direction over a certain period of time.

2. How do you calculate the x components of average acceleration?

To calculate the x components of average acceleration, you divide the change in velocity in the x direction by the change in time.

3. What is the unit of measurement for x components of average acceleration?

The unit of measurement for x components of average acceleration is meters per second squared (m/s^2).

4. How do x components of average acceleration relate to the overall average acceleration?

The x components of average acceleration are one part of the overall average acceleration, which also includes the y and z components. The overall average acceleration can be found by taking the square root of the sum of the squares of all three components.

5. What factors can affect the x components of average acceleration?

The x components of average acceleration can be affected by factors such as friction, air resistance, and external forces acting on the object in the horizontal direction.

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